What is the molecular weight they determined for this compound ?

The freezing point of water, H2O, is 0.000 °C at 1 atmosphere. Kf(water) = -1.86 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 13.30 grams of the compound were dissolved in 250.1 grams of water, the solution began to freeze at -1.593 °C. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight they determined for this compound ?

1 Answer
anor277
May 18, 2018

#"Molecular mass"~=60*g*mol^-1#

Explanation:

The freezing point depression observed for a solution, is proportional to the molality of the solute in solution....and here we have the appropriate parameters...

#underbrace(DeltaT_f)_"freezing point depression"=mxxk_f#...

..where #k_f=1.86*K*mol^-1#...and #m="molality"...#

#m=(DeltaT_f)/(1.86*K*mol^-1)=(1.593*K)/(1.86*K*mol^-1)=0.856*mol*kg^-1#...

And so #m=0.856*mol*kg^-1=((13.30*g)/("molar mass"))/(0.2501*kg)#...

#"molar mass"=(13.30*g)/(0.2501*kg)xx1/(0.856*mol*kg^-1)#

#=??*g*mol^-1#

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